Re: NegativeArraySizeException ... IndexOutOfBoundsException ...

From:

Lew <lew@lewscanon.com>

Newsgroups:

comp.lang.java.programmer

Date:

Tue, 12 Jan 2010 11:12:37 -0800 (PST)

Message-ID:

<dbe988bf-ad61-48f9-b189-a140d7d429a9@g25g2000yqd.googlegroups.com>

Lew wrote:

Where will you store the array? Either you have a crapload of RAM
(one bit per atom storage density?) or the largest-capacity storage
device ever invented (one bit per atom storage density?).

What's the average retrieval latency? Even at one bit per atom
storage density, it must take even a light beam noticeable time to
reach the further reaches of the storage device; anything slower like
a semiconductor must take a really long time.

A silicon crystal lattice has a lattice spacing of just over half a
nanometer, or 5.4 x 10^-10 m. A three-dimensional storage medium for
a 9 x 10^18-element array would hold juar over 2 x 10^6 elements to
the side. An average access would be halfway in each dimension, or
10^6 elements, which in a silicon lattice is about 5.4 x 10-4 m, times
three for a total traversal distance of about 1.6 x 10^-3 m. Each
way. For a round trip slightly over 3 x 10^-3 m. A light beam
travels that in 0.1 microseconds (10^-7 s).

Drat! Mixed up my CGS and MKS. That's 10^-5 s, or 10 microseconds.

That's about 20,000

clock cycles of latency on a modern processor,
far more on the future processors of 2074.

Either we'll find a sparse representation for such arrays, we'll
invent much denser storage media and better ways to access them, we'll
find some way to keep the processor busy during that latency, or we'll
use super-luminal access speeds, perhaps through quantum
superposition.

--
Lew